Home
Class 12
MATHS
If p is the length of the perpendicular ...

If p is the length of the perpendicular from a focus upon the tangent at any point P of the the ellipse `x^2/a^2+y^2/b^2=1` and r is the distance of P from the focus , then `(2a)/r-(b^2)/(p^2)` is equal to

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( \frac{2a}{r} - \frac{b^2}{p^2} \) where \( p \) is the length of the perpendicular from the focus to the tangent at point \( P \) on the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), and \( r \) is the distance from the focus to point \( P \). ### Step-by-Step Solution: 1. **Identify the Point on the Ellipse**: Let the point \( P \) on the ellipse be represented in parametric form as: \[ P(a \cos \theta, b \sin \theta) \] 2. **Equation of the Tangent**: The equation of the tangent to the ellipse at point \( P \) is given by: \[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \] 3. **Focus of the Ellipse**: The foci of the ellipse are located at \( (ae, 0) \) and \( (-ae, 0) \), where \( e = \sqrt{1 - \frac{b^2}{a^2}} \) is the eccentricity of the ellipse. 4. **Length of the Perpendicular from the Focus**: The length \( p \) of the perpendicular from the focus \( (ae, 0) \) to the tangent line can be calculated using the formula for the distance from a point to a line: \[ p = \frac{|a e \cos \theta + b \sin \theta - 1|}{\sqrt{\left(\frac{1}{a \cos \theta}\right)^2 + \left(\frac{1}{b \sin \theta}\right)^2}} \] Simplifying this expression leads to: \[ p = \frac{ab(e \cos \theta - 1)}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}} \] 5. **Distance from the Focus to Point P**: The distance \( r \) from the focus \( (ae, 0) \) to the point \( P(a \cos \theta, b \sin \theta) \) is given by: \[ r = \sqrt{(a e - a \cos \theta)^2 + (b \sin \theta)^2} \] Simplifying this expression gives: \[ r = a \sqrt{(e - \cos \theta)^2 + \frac{b^2}{a^2} \sin^2 \theta} \] 6. **Substituting \( p^2 \) and \( r \)**: Now, we need to find \( \frac{b^2}{p^2} \) and substitute it into the expression \( \frac{2a}{r} - \frac{b^2}{p^2} \). 7. **Final Expression**: After substituting and simplifying, we find: \[ \frac{2a}{r} - \frac{b^2}{p^2} = 1 \] ### Conclusion: Thus, the final answer is: \[ \frac{2a}{r} - \frac{b^2}{p^2} = 1 \]
Promotional Banner

Topper's Solved these Questions

  • ELLIPSE

    ARIHANT MATHS ENGLISH|Exercise Ellipse Exercise 5: Matching Type Questions|3 Videos
  • ELLIPSE

    ARIHANT MATHS ENGLISH|Exercise Exercise (Statement I And Ii Type Questions)|8 Videos
  • ELLIPSE

    ARIHANT MATHS ENGLISH|Exercise Exercise (Passage Based Questions)|15 Videos
  • DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|7 Videos
  • ESSENTIAL MATHEMATICAL TOOLS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Single Integer Answer Type Questions)|3 Videos

Similar Questions

Explore conceptually related problems

Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse x^2/a^2+y^2/b^2=1 is b^2

Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse x^2/a^2+y^2/b^2=1 is b^2

The locus of the foot of the perpendicular from the foci an any tangent to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , is

Find the locus of the foot of the perpendicular drawn from the center upon any tangent to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1.

If P is the length of perpendicluar drawn from the origin to any normal to the ellipse x^(2)/25+y^(2)/16=1 , then the maximum value of p is

If the normal to the ellipse x^2/25+y^2/1=1 is at a distance p from the origin then the maximum value of p is

The distance from the foci of P (a, b) on the ellipse x^2/9+y^2/25=1 are

If the tangent to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 makes intercepts p and q on the coordinate axes, then a^(2)/p^(2) + b^(2)/q^(2) =

The distance of the point 'theta' on the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 from a focus, is

if P is the length of perpendicular from origin to the line x/a+y/b=1 then prove that 1/(a^2)+1/(b^2)=1/(p^2)